Let $\{a_{m,j}\}_{m,j \in \Bbb N} \in \Bbb R_+$ such that for each $m$ $$b_m=\sum_{j=1}^\infty a_{m,j}<\infty$$ and $$\lim_{j \to \infty} \frac{a_{m+1,j}}{a_{m,j}}=0$$
I would like to know if is true that:
Exist $k \in \Bbb N$ such that $b_{m+1}
Asymptotic analysis of a double sequence
1
$\begingroup$
sequences-and-series
analysis
asymptotics
1 Answers
1
No. Take for instance $a_{1,j} = \frac{1}{j^2}$ for all $j\geq 1$, and, for $m\geq 1$, $$ a_{m+1,j} = \begin{cases} \frac{a_{m,j}}{j} & \text{ if } j\geq 2\\ b_m+1 & \text{ if } j=1 \end{cases} $$ Then your assumptions are satisfied, yet $b_{m+1} > a_{m+1,1} = b_m+1 > b_m$ for all $m$.
The issue is that the assumption that $\lim_{j \to \infty} \frac{a_{m+1,j}}{a_{m,j}}=0$ is only an asymptotic guarantee, and does not tell you anything about the first terms — which can do whatever they want.